Using Euclidean geometry to prove that two lengths are equal The incircle of triangle ABC is tangent to BC, CA and AB at D, E and F, respectively.
Suppose the incircle intersects again with AD at a point X such that AX = XD. Also
XB and XC intersect again with the incircle at points Y and Z, respectively. Show that
EY = F Z.
I noted that it should be true that FE = YZ = AX = XD. If this is proven to be true, then we can certainly get that FZ = EY. The problem is, I haven't been able to show that.
Another thing I thought about was line chasing. If I was able to show that FY = EZ, we are done.
By letting BD = x, AF = z, and EC = y, we can find the length of AD = 2AX = 2XD using Stewart's theorem. We could also do the same for FD, then FX. To find FY using Stewart's theorem, however, I need the lengths of BY and YX. We can get that BY $\times$ BX = $x^2$ by power of a point, but I don't think it's possible to explicitly find FY. If we can find FY, then the proof for EZ should be analogous.
Also, it would be nice if solutions or tips relate to euclidean geometry, as that is what I am specifically trying to learn.
 A: $FE=YZ=XD=AD$ does not follow from problem condition.
Consider line $a$ going through $F$ and parallel to $XD$. Let this line intersect with incircle in point $Y_1$ and intersect $BC$ in point $G$.
From circle: $AF^2=AD\cdot XD=2XD^2\Rightarrow AF=XD\sqrt{2}$
$BA=AF+BF=XD\sqrt{2}+BD$
From triangles similarity: $BG/BD=BF/BA\Rightarrow BG=BD\cdot BF/BA=\frac{BD^2}{XD\sqrt{2}+BD}$
$GD=BD-BG=BD-\frac{BD^2}{XD\sqrt{2}+BD}=\frac{\sqrt{2}\ XD\cdot BD}{XD \sqrt{2} + BD}$
From triangle similarity: $GF/AD=BF/BA\Rightarrow GF=AD\cdot BF/BA=\frac{2XD\cdot BD}{XD\sqrt{2}+BD}$
From circle: $GF\cdot GY_1=GD^2 \Rightarrow GY_1=\frac{GD^2}{GF}=\frac{2 XD^2\cdot BD^2}{(XD\sqrt{2}+BD)^2}\frac{XD\sqrt{2}+BD}{2XD\cdot BD}=\frac{XD\cdot BD}{XD\sqrt{2}+BD}$
$GY_1/BG=XD/BD\Rightarrow$ triangles $BGY_1$ and $BDX$ are similar, therefore $\angle DBX=\angle GBY_1$, therefore $X$, $Y_1$ and $B$ are collinear, therefore $Y_1=Y$. That means that $FY$ is parallel to $XD$.
In analogous way: $EZ$ is parallel to $XD$, therefore $EZ$ parallel to $FY$, then $EZYF$ is isosceles trapezoid, which diagonals are equal.
